Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Determine mean and standard deviation of first n terms of an A.P. whose first term is a and common difference is d.

Answers (1)

The first n terms of an A.P are given whose first term is a and common difference is d. We have to find mean and standard deviation.

The given AP in tabular form is as shown below,

\begin{array}{|l|l|l|} \hline \multicolumn{1}{|c|} {x_{i}} & \multicolumn{1}{|c|} {d_{i}=x_{i}-a} & d_{i}^{2} \\ \hline a & 0 & 0 \\ \hline a+d & d & d^{2} \\ \hline a+2 d & 2 d & 4 d^{2} \\ \hline a+3 d & 3 d & 9 d^{2} \\ \hline- & - & - \\ \hline a+(n-1) d & (n-1) d & (n-1)^{2} d^{2} \\ \hline \end{array}
\\ \text{Here we have assumed a as mean } \\ \text{~ Given that AP has n terms.} \\ \text{ And we know the sum of all terms of AP can be written as }

\\ \sum x_{i}=\frac{n}{2} \left[ 2a+ \left( n-1 \right) d \right] \\ \overline{x}=\frac{ \sum x_{i}}{n}=\frac{2a+ \left( n-1 \right) d}{2}=a+\frac{ \left( n-1 \right) }{2}d \\ \sum d_{i}= \sum \left( x_{i}-a \right) =d \left[ 1+2+3+ \ldots + \left( n-1 \right) \right] =d \left( \frac{n \left( n-1 \right) }{2} \right)


\\ \sum d_{i}^{2}= \sum \left( x_{i}-a \right) ^{2}=d^{2} \left[ 1^{2}+2^{2}+3^{2}+ \ldots + \left( n-1 \right) ^{2} \right] =d^{2} \left( \frac{n \left( n-1 \right) \left( 2n-1 \right) }{6} \right) \\\\ \\ \sigma =\sqrt {\frac{ \sum \left( x_{i}-a \right) ^{2}~}{n}- \left( \frac{ \sum \left( x_{i}-a \right) }{n} \right) ^{2}} \\\\ \\ =\sqrt {\frac{d^{2} \left( \frac{n \left( n-1 \right) \left( 2n-1 \right) }{6} \right) }{n}-\frac{d^{2}n^{2} \left( n-1 \right) ^{2}}{4n^{2}}} \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) \left( 2n-1 \right) }{6}-\frac{d^{2} \left( n-1 \right) ^{2}}{4}} \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{2n-1}{3}-\frac{n-1}{2} \right) } \\\\ \\ =\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{4n-2-3n+3}{6} \right) }=\sqrt {\frac{d^{2} \left( n-1 \right) }{2} \left( \frac{n+1}{6} \right) }=d\sqrt {\frac{n^{2}-1}{12}} \\\\

 

 

Posted by

infoexpert21

View full answer

Similar Questions

Latest Question